Question

方程根号下2(x-1)^2+2(y-1)^2=x+y-2表示的曲线是

Answer 1

√{2(x-1)^2+2(y-1)^2} = x+y-2

x+y-2≥0,x+y≥2

原式化为：√{2(x-1)^2+2(y-1)^2} = (x-1)+(y-1)

两边平方：2(x-1)^2+2(y-1)^2 = (x-1)^2+(y-1)^2+2(x-1)(y-1)

化简：

(x-1)^2+(y-1)^2 - 2(x-1)(y-1) = 0

(x-1-y+1)^2 = 0

x-y=0

y=x

又：x+y≥2

∴表示射线y=x,定义域x∈【1,+∞）